Squeeze Theorem/Functions/Proof 2
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Theorem
Let $a$ be a point on an open real interval $I$.
Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.
Suppose that:
- $\forall x \ne a \in I: g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
- $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right) = L$
Then:
- $\displaystyle \lim_{x \mathop \to a} \ f \left({x}\right) = L$
Proof
Let $f, g, h$ be real functions defined on an open interval $\left({a \,.\,.\, b}\right)$, except possibly at the point $c \in \left({a \,.\,.\, b}\right)$.
Let:
- $\displaystyle \lim_{x \mathop \to c} \ g \left({x}\right) = L$
- $\displaystyle \lim_{x \mathop \to c} \ h \left({x}\right) = L$
- $g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ except perhaps at $x = c$.
Let $\left \langle {x_n} \right \rangle$ be a sequence of points of $\left({a \,.\,.\, b}\right)$ such that:
- $\forall n \in \N_{>0}: x_n \ne c$
and:
- $\displaystyle \lim_{n \mathop \to \infty} \ x_n = c$
By Limit of Function by Convergent Sequences:
- $\displaystyle \lim_{n \mathop \to \infty} \ g \left({x_n}\right) = L$
and:
- $\displaystyle \lim_{n \mathop \to \infty} \ h \left({x_n}\right) = L$
Since:
- $g \left({x_n}\right) \le f \left({x_n}\right) \le h \left({x_n}\right)$
it follows from the Squeeze Theorem for Real Sequences that:
- $\displaystyle \lim_{n \mathop \to \infty} \ f \left({x_n}\right) = L$
The result follows from Limit of Function by Convergent Sequences.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): Appendix: $\S 18.6$: Limits of functions